Inheriting Multiple Classes in TypeScript

Question

In what ways can you get around the problem only in order to allow the extension of no more than one other class.

class Bar { doBarThings() { //... } } class Bazz { doBazzThings() { //... } } class Foo extends Bar, Bazz { doBarThings() { super.doBarThings(); //... } } 

This is currently not possible, TypeScript will throw an error. This problem can be solved in other languages ​​using interfaces, but the solution to the problem with them is not possible in TypeScript.

Suggestions are welcome!

Answer 1

This is possible with the interfaces:

 interface IBar { doBarThings(); } interface IBazz { doBazzThings(); } class Foo implements IBar, IBazz { doBarThings() {} doBazzThings(){} } 

But if you want to implement this in super / base style, then you will need to do something else, for example:

 class FooBase implements IBar, IBazz{ doBarThings() {} doBazzThings(){} } class Foo extends FooBase { doFooThings(){ super.doBarThings(); super.doBazzThings(); } } 

Answer 2

This is my way of extending multiple classes. This allows for pretty nice type safety. I have not yet been able to find any significant flaws in this approach, which works the same as with multiple inheritance.

 // tslint:disable: max-classes-per-file class Base {} type Constructor<I = {}> = new (...args: any[]) => I | {}; interface IBar { doBarThings(): void; } interface IBazz { doBazzThings(): void; } function Bar<T extends Constructor>(constructor: T) { return class extends constructor implements IBar { public doBarThings() { console.log("Do bar!"); } }; } function Bazz<T extends Constructor>(constructor: T) { return class extends constructor implements IBazz { public doBazzThings() { console.log("Do bazz!"); } }; } class Foo extends Bar(Bazz(Base)) implements IBar, IBazz { public doBarThings() { super.doBarThings(); console.log("Override mixin"); } } const foo = new Foo(); foo.doBazzThings(); // Do bazz! foo.doBarThings(); // Do bar! // Override mixin